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48 SYSTEM MODELLING
qi
h1 |
R1 |
R2 |
h2 |
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qo |
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A1, ρ, g |
P1 |
q1 |
P2 A2, ρ, g |
P3 |
P4 |
Figure 2.24 Two tank liquid level system
the second tank at the rate of qo through an orifice of R2. Derive the mathematical model for the system.
Solution
The solution is similar to Example 2.12, but we have to consider both tanks. For tank 1,
qi − q1 = |
A1 d p |
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ρg dt |
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or |
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qi |
A1 d p |
(2.100) |
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= |
+ q1. |
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ρg |
dt |
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But |
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thus (2.100) becomes |
p = hρg, |
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qi |
d h1 |
(2.101) |
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= A1 |
+ q1 |
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dt |
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Since |
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or |
p1 − p2 = R1q1 |
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q |
p1 − p2 |
h1ρg − h2ρg |
, |
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1 |
= |
= |
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R1 |
R1 |
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we have |
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qi = A1 |
d h1 |
+ |
ρgh1 |
− |
ρgh2 |
. |
(2.102) |
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dt |
R1 |
R1 |
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For tank 2, |
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q1 − q0 = |
A2 d p |
(2.103) |
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, |
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ρg |
dt |
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and with
p = hρg
THERMAL SYSTEMS |
49 |
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(2.103) becomes |
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q1 − qo = A2 |
d h2 |
(2.104) |
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. |
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dt |
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But |
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q |
1 |
= |
p1 − p2 |
and |
q |
o = |
p2 − p3 |
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R2 |
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R1 |
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so |
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q |
1 |
− |
q |
0 |
= |
p1 − p2 |
− |
p2 − p3 |
= |
h1ρg − h2ρg |
− |
h2ρg |
. |
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R1 |
R2 |
R1 |
R2 |
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Substituting in (2.104), we obtain
d h2 |
ρgh1 |
1 |
1 |
ρgh2 = 0. |
(2.105) |
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A2 |
− |
+ |
+ |
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dt |
R1 |
R1 |
R2 |
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Equations (2.102) and (2.105) describe the behaviour of the system. These two equations can be represented in matrix form as
0 |
A2 |
˙ |
2 |
+ |
ρg/ R1 |
ρg/ R1 |
ρg/ R2 |
h2 |
= |
0 |
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h˙ |
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A1 |
0 |
h1 |
ρg/ R1 |
−ρg/ R1 |
h1 |
qi . |
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− |
+ |
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2.5 THERMAL SYSTEMS
Thermal systems are encountered in chemical processes, heating, cooling and air conditioning systems, power plants, etc. Thermal systems have two basic components: thermal resistance and thermal capacitance. Thermal resistance is similar to the resistance in electrical circuits. Similarly, thermal capacitance is similar to the capacitance in electrical circuits. The across variable, which is measured across an element, is the temperature, and the through variable is the heat flow rate. In thermal systems there is no concept of inductance or inertance. Also, the product of the across variable and the through variable is not equal to power. The mathematical modelling of thermal systems is usually complex because of the complex distribution of the temperature. Simple approximate models can, however, be derived for the systems commonly used in practice.
Thermal resistance, R, is the resistance offered to the heat flow, and is defined as:
R |
= |
T2 − T1 |
, |
(2.106) |
q |
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where T1 and T2 are the temperatures, and q is the heat flow rate.
Thermal capacitance is a measure of the energy storage in a thermal system. If q1 is the heat flowing into a body and q2 is the heat flowing out then the difference q1 − q2 is stored by the
body, and we can write |
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q1 − q2 = mc |
d T |
(2.107) |
|
, |
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dt |
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50 SYSTEM MODELLING
Tw |
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Tr |
To |
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Heater |
q |
qwo |
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qrw |
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room |
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wall |
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ambient |
Figure 2.25 Simple thermal system
where m is the mass and c is the specific heat capacity of the body. If we let the heat capacity be denoted by C , then
d T |
(2.108) |
q1 − q2 = C dt , |
where C = mc.
An example thermal system model is given below.
Example 2.14
Figure 2.25 shows a room heated with an electric heater. The inside of the room is at temperature Tr and the walls are assumed to be at temperature Tw . If the outside temperature is To , develop a model of the system to show the relationship between the supplied heat q and the room temperature Tr .
Solution
The heat flow from inside the room to the walls is given by
qr w = Tr − Tw , Rr
where Rr is the thermal resistance of the room.
Similarly, the heat flow from the walls to the outside is given by
qwo = Tw − To , Rw
(2.109)
(2.110)
where Rw is the thermal resistance of the walls. Using (2.108) and (2.109), we can write
− |
Rr |
= |
1 |
dt |
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q |
Tr − Tw |
C |
d Tr |
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or |
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˙ |
Tr |
Tw |
(2.111) |
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C1 Tr + |
− |
= q. |
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Rr |
Rr |
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THERMAL SYSTEMS |
51 |
Also, using (2.108) and (2.110), we can write
Rr |
− |
Rw |
= |
2 dt |
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Tr − Tw |
Tw − To |
C |
d Tw |
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or
˙ |
Tr |
+ |
1 |
+ |
1 |
Tw = |
To |
(2.112) |
|
C2 Tw − |
Rr |
Rr |
Rw |
Rw . |
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Equations (2.111) and (2.112) describe the behaviour of the system and they can be written in matrix form as
0 C2 |
˙ |
+ |
1/ Rr |
1/ Rr |
1/ Rw |
Tw |
= |
To / Rw |
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T˙w |
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C1 |
0 |
Tr |
1/ Rr |
−1/ Rr |
Tr |
q |
. |
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− |
+ |
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Example 2.15
Figure 2.26 shows a heated stirred tank thermal system. Liquid enters the tank at the temperature Ti with a flow rate of W . The water is heated inside the tank to temperature T . The temperature leaves the tank at the same flow rate of W . Derive a mathematical model for the system, assuming that there is no heat loss from the tank.
Solution
The following equation can be written for the conservation of energy:
Q p + Qi = Ql + Qo , |
(2.113) |
where Q p is the heat supplied by the heater, Qi is the heat flow via the liquid entering the tank, Ql is the heat flow into the liquid and Qo is the heat flow via the liquid leaving the tank.
Now,
Qi = WC p Ti |
(2.114) |
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where W is the flow rate (kg/s), and C p is the specific heat capacity of the liquid. Also |
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Qo = WC p T |
(2.115) |
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and |
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dT |
(2.116) |
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Ql = C |
, |
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dt |
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mixer
Ti
W
heater Qp
T
W, T
Figure 2.26 Heated stirred tank for Example 2.15